Let $R$ be a Noetherian ring and $I$ an ideal of $R$. If $n$ is the cohomological dimension of $I$, then why is the following isomorphism true: $$H_{I}^{n}(M)\cong H_{I}^{n}(R)\otimes_R M.$$

The cohomological dimension of $I$ is defined to be the supremum of the set of integers $i$ such that $H_{I}^{i}(M)\neq 0$ for some $R$-module $M$.